Find the volume of the solid obtained by rotating the region enclosed by y=x^3, y=25x, x ≥ 0

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To find the volume of the solid obtained by rotating the region enclosed by the curves y = x^3, y = 25x, and x ≥ 0, we can use the method of cylindrical shells.

Here's how to do it:

1. First, let's find the points of intersection between the two curves. Set them equal to each other and solve for x:
x^3 = 25x

Rearrange the equation:
x^3 - 25x = 0

Factor out an x:
x(x^2 - 25) = 0

We get two solutions:
x = 0 and x = ±5

Since we are only interested in the region with x ≥ 0, we discard the negative solution x = -5.

2. Next, let's find the height of the cylindrical shell at any given value of x. The height is the difference between the two curves: y = 25x - y = x^3.
So, the height h(x) = (25x) - (x^3) = 25x - x^3.

3. The radius of the cylindrical shell is simply the x-coordinate. Therefore, the radius r(x) = x.

4. Now, we can set up the integral to find the volume. The volume of a cylindrical shell is given by the formula: V = ∫2πrh(x)dx.

The limits of integration for x are from 0 to 5 (since we are only interested in the region where x ≥ 0).

The volume is then given by:
V = ∫[0 to 5] (2πx)(25x - x^3) dx.

5. Evaluate the integral:
V = 2π ∫[0 to 5] (25x^2 - x^4) dx.

Integrate each term separately:
V = 2π [ (25/3)x^3 - (1/5)x^5 ] evaluated from 0 to 5.

Plugging in the limits:
V = 2π [ (25/3)(5^3) - (1/5)(5^5) ].

Simplify the expression:
V = 2π [ (125/3)x^3 - (3125/5)x^5 ].

Finally, calculate the volume using a calculator or numerical approximation.

And that's how you find the volume of the solid obtained by rotating the region enclosed by y = x^3, y = 25x, and x ≥ 0 using the method of cylindrical shells.